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Calculate pH values of (i) 0.2 M `H_(2)SO_(4)` solution (ii) 0.2M `Ca(OH)_(2)` solution. |
Answer» (i) pH, value of 0.2M `H_(2)SO_(4)` solution `H_(2)SO_(4)` is a dibasic acid and dissociates in solution as : `H_(2)SO_(4) hArr 2H^(+) + SO_(4)^(2-)` Concentration of `[H^(+)] = 2 xx 0.2 = 0.4 M` pH of solution `=- log [H_(3)O^(+)]=-log[0.4]=- (-0.3979)= 0.3979` (ii) pH value of 0.2M `Ca(OH)_(2)` solution `Ca(OH)_(2)` is a diacid base and dissociates in solution as : `Ca(OH)_(2) hArr Ca^(2+) + 2OH^(-)` Concentration of `[OH^(-)] = 2 xx 0.2 =0.4M` `pOH=- log (OH^(-)) =- log [0.4] =- (-0.3979) =0.3979` `pH = 14 - pOH = 14 - 0.3979 = 13. 6021` |
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